Question: Let $x^4+y^2=17$. What is the value of $\dfrac{d^2y}{dx^2}$ at the point $(-2,1)$ ? Give an exact number.
Solution: Notice that the equation defines $y$ implicitly—we don't have an explicit expression for $y$ in terms of $x$. So we will have to use implicit differentiation. If we differentiate the equation once, we will be able to get an expression for $\dfrac{dy}{dx}$. Then we can differentiate the equation again to get an expression for $\dfrac{d^2y}{dx^2}$. Let's start by finding $\dfrac{dy}{dx}$. $\dfrac{dy}{dx}=-\dfrac{2x^3}{y}$ Now we can differentiate $\dfrac{dy}{dx}$ to find $\dfrac{d^2y}{dx^2}$. $\dfrac{d^2y}{dx^2}=-\dfrac{6x^2y^2+4x^6}{y^3}$ Finally, let's plug ${x=-2}$ and ${y=1}$ into the expression we got: $\begin{aligned} \left.-\dfrac{6 x^2 y^2+4 x^6}{ y^3}\right\rvert_{({-2},{1})}&=-\dfrac{6({-2})^2({1})^2+4({-2})^6}{({1})^3} \\\\ &=-280 \end{aligned}$ In conclusion, the value of $\dfrac{d^2y}{dx^2}$ at the point $(-2,1)$ is $-280$.